1. Field of the Invention
The present invention generally relates to an wireless communications system. More particularly, the present invention relates to a method and apparatus for reducing other cell interference by using a Base Station (BS)-specific orthogonal code or quasi-orthogonal code.
2. Description of the Related Art
Beginning with analog cellular service called 1st Generation (1G), mobile communication systems are evolving toward 4th Generation (4G) that provides ultra high-speed multimedia service beyond 2nd Generation (2G) digital technology and 3rd Generation (3G) providing high-speed multimedia service in International Mobile Telecommunications-2000 (IMT-2000). The 4G mobile communication system is designed to support higher data rates, aiming at data transmission at or above 100 MBps. Under a multipath radio channel environment, the 4G mobile communication system compensates for multipath fading and ensures transmission of burst packet data that has rapidly increased in amount along with provisioning of packet service.
OFDMA is a promising candidate for radio access technology satisfying 4G mobile communication requirements. OFDMA is a special case of MultiCarrier Modulation (MCM) in which input data is converted to as many parallel data sequences as subcarriers are used and the input data is modulated in the subcarriers, prior to transmission.
Since the number of subcarriers changes according to a user-requested data rate, OFDMA enables efficient resource distribution and increases transmission efficiency. Due to its feasibility for multiple subcarriers (i.e. a large Fast Fourier Transform (FFT) size), OFDMA is efficient for a wireless communication system having cells with a relatively long time delay spread.
FIG. 1 is a block diagram of a typical OFDMA transmission structure. In the illustrated case of FIG. 1, transmitters 100, 120 and 140 are provided in a plurality of BSs controlling their respective cells, BS0, BS1 and BS2.
Referring to FIG. 1, Forward Error Correction (FEC) encoders 102, 122 and 142 encode information sequences as source data to be transmitted from their BSs, i.e. BS0, BS1 and BS2. Symbol mappers 104, 124 and 144 modulate the coded data in Quadrature Phase Shift Keying (QPSK)/16-ary Quadrature Amplitude Modulation (16QAM)/64-ary QAM (64QAM), thus producing modulation symbols s0(k), s1(k) and s2(k). Repeaters 106, 126, and 146 repeat the modulation symbols s0(k), s1(k) and s2(k) according to repetition numbers set in their BSs (BS0, BS1 and BS2) and allocate the repeated symbols to a plurality of subcarriers.
Scramblers 108, 128 and 148 scramble the signal sequences SR,0(k), SR,1(k) and SR,2(k) allocated to the subcarriers with BS-specific scrambling sequences sc0(k), sc1(k) and sc2(k). Inverse Fast Fourier Transform (IFFT) processors 110, 130 and 150 convert the scrambled signals to time-domain signals. The time-domain signals are sent to users within the cell areas of BS0, BS1 and BS2 via antennas 112, 132 and 152.
The operation of transmitters 100, 120 and 140 is formulated as follows.
The outputs of symbol mappers 104, 124 and 144 are given as Equation (1),s(m), m=0, . . . , M−1  (1)where m denotes a bit index and M denotes the length of a modulation symbol, i.e. the number of bits per modulation symbol.
Repeaters 106, 126 and 146 repeat s(m) R times as expressed by Equation (2). Thus,sR(k)=s(k mod M), k=0, . . . ,N−1  (2)where k denotes a subcarrier index and mod represents a modulo operation. N is the total length of a symbol which is repeated R times as expressed by Equation (3). Therefore,N=RM  (3)
Let a scrambling sequence with 1s or −1s be denoted by sc(k). Then, the outputs of the scramblers 108, 128 and 148 are expressed as Equation (4),x(k)=sc(k)sR(k), k=0, . . . , N−1  (4)
A receiver of a Mobile Terminal (MT) receives interference signals from neighbor cells, i.e. neighbor cell interference signals as well as a signal from a serving BS. If BS0 is a serving cell and BS1 is a neighbor cell in FIG. 1, the received signal y(k) includes noise n(k), expressed as Equation (5),y(k)=h0(k)x0(k)+h1(k)x1(k)+n(k)  (5)where x0(k) and x1(k) denote signals transmitted from BS0 and BS1, h0(k) denotes a channel frequency response representing a kth subchannel between BS0 and the MT, and h1(k) denotes a channel frequency response between BS1 and the MT.
FIG. 2 is a block diagram of a typical OFDMA receiver.
Referring to FIG. 2, an FFT processor 202 converts a time-domain signal y(k) received through an antenna 200 to a frequency-domain signal by FFT and a descrambler 204 descrambles the FFT signals as modeled by Equation (6),zp(k)=sc0(k)y(k)  (6)where sc0(k) denotes a scrambling sequence allocated to the serving cell (e.g. BS0). A channel compensator 208 compensates the descrambled signal using a channel frequency response estimated by a channel estimator 206 as expressed by Equation (7).zR(k)=zp(k)=sc0(k)y(k)/h0(k)  (7)where h0(k) denotes the channel frequency response estimated for the serving cell BS0.
A combiner 210 accumulates the channel-compensated signal as many times as the repetition number R of the repeater 106 as given by Equation (8).
                                                                                          z                  ⁡                                      (                    m                    )                                                  =                                                      1                                                                                                              ⁢                      R                                                        ⁢                                                            ∑                                              r                        ⁢                                                                                                  =                                                                                                  ⁢                        0                                                                                                                                              ⁢                                                  R                          ⁢                                                                                                          -                                                                                                          ⁢                          1                                                                                      ⁢                                                                  z                                                                                                                                  ⁢                          R                                                                    ⁡                                              (                                                  r                          +                          mR                                                )                                                                                                                                                                    =                                                                            1                                                                                                                        ⁢                        R                                                              ⁢                                                                  ∑                                                  r                          ⁢                                                                                                          =                                                                                                          ⁢                          0                                                                                                                                                          ⁢                                                      R                            ⁢                                                                                                                  -                                                                                                                  ⁢                            1                                                                                              ⁢                                                                        s                                                                                                                                            ⁢                                                          R                              ,                                                                                                                          ⁢                              0                                                                                                      ⁢                                                  (                                                      r                            +                            mR                                                    )                                                                                                      +                                                            1                                                                                                                        ⁢                        R                                                              ⁢                                                                  ∑                                                  r                          ⁢                                                                                                          =                                                                                                          ⁢                          0                                                                                                                                                          ⁢                                                      R                            ⁢                                                                                                                  -                                                                                                                  ⁢                            1                                                                                              ⁢                                              i                        ⁡                                                  (                                                      r                            +                            mR                                                    )                                                                                                                                                        ⁢                                  ⁢                              i            ⁢                          (              k              )                                =                                                    sc                                                                                          ⁢                  0                                            ⁡                              (                k                )                                      ⁢                                          (                                                                            h                                                                                                                        ⁢                        1                                                              ⁢                                          (                      k                      )                                        ⁢                                          x                                                                                                                        ⁢                        1                                                              ⁢                                          (                      k                      )                                                        +                                      n                    ⁢                                          (                      k                      )                                                                      )                            /                                                h                                                                                                    ⁢                    0                                                  ⁡                                  (                  k                  )                                                                                        (        8        )            where SR,0(k) denotes a repeated transmission signal from BS0 and I(k) denotes a noise and interference signal.
Considering sR,0(0+mR)=sR,0(1+mR)= . . . =sR,0(R−1+mR), then z (m) is given by Equation (9) as,
                              z          ⁡                      (            m            )                          =                                            s              0                        ⁡                          (              m              )                                +                                    1              R                        ⁢                                          ∑                                  r                  =                  0                                                  R                  -                  1                                            ⁢                              i                ⁡                                  (                                      m                    +                    Mr                                    )                                                                                        (        9        )            where s0(m) denotes a modulation symbol from BS0. Assuming that the noise and interference signal is Additive White Gaussian Noise (AWGN), an averaging effect reduces the noise and interference signal, thus improving the reception performance of the MT.
Wireless Broadband (WiBro), which is an Institute of Electrical and Electronics Engineers (IEEE) 802.16-based OFDMA system, basically uses a frequency reuse factor of 1. With the frequency reuse factor of 1, the WiBro system is advantageous in terms of frequency efficiency but suffers from interference between BSs because all subcarriers used in one BS are overlapped with those of neighbor BSs. The neighbor BS interference signals degrade the reception performance of an MT at a cell boundary and often interrupt communications during handover.
FIG. 3 illustrates a typical MT located in an overlap region among three cells. Three BSs 312, 314 and 316 cover cell areas 302, 304 and 306, respectively and an MT 310 is located equidistantly from BSs 312, 314 and 316. In the case where the MT is located at any place within the overlap area as well as at the equidistant area, signals from BSs other than a serving BS selected by MT 310 interfere with a signal from the serving BS. To avert the interference problem at the cell boundary, the IEEE 802.16 standards provide that a low modulation order like QPSK, a low FEC coding rate, and a repetition number of up to 6 are used for a transmission signal from a BS. Despite these efforts, an existing MT receiver has a high outage probability in the vicinity of the cell boundary and experiences degradation of handover performance on a fading channel. To fundamentally solve this problem, the frequency reuse factor should be at least 3, but at the expense of decreasing frequency efficiency to ⅓ and cell planning complexity.